Sources page probability recreations

Theorema CXIIII, pp. 75-77. M-(9, 11) starting D = 400 apart. Then does the problem in general, giving a diagram which I can't follow.

Theorema CXV, pp. 77-78. MR-(10, 15; 100), then MR-(a, b; D) in general.

Wingate/Kersey. 1678?.

Quest. 14, pp. 485-486. O-(14, 22), T = 8.

Quest. 15, pp. 486-487. O-(9, 7) on a circular island of circumference 36.

Quest. 16, p. 487. O-(8, 0; 1, 1).

Quest. 17, pp. 487-488. M-(8, 6; 140).

Quest. 18, pp. 488-489. M-(a, a + 5/2; 100) takes time 8.

Quest. 19, pp. 489-490. M-(11/2, 17/3; 134) going opposite ways round a circular island.

Quest. 23, pp. 491-492. Hare is 100 hare-leaps ahead of a hound. Hare takes five leaps while the hound takes four, but three hound leaps equal four hare leaps.

Edward Cocker. Arithmetic. Op. cit. in 7.R. 1678. Chap. 10, quest. 32. 1678, p. 181; 1715: p. 121; 1787: p. 106. This problem is attributed to Moor's or More's Arith. cap. 8, qu. 7. 1678 states O-(40, 50), T = 3, giving answers 12 days and 600 miles. 1715 states O-(40, 50), T = 3, giving answers 32 days and 600 miles. 1787 states O-(48, 50), T = 3, giving answers 12 days and 600 miles. I am surprised at the misprints here.

Wells. 1698.

No. 109, p. 207. Meeting: MR-(1, 3; 50).

No. 116, p. 208. Overtaking: O-(6, 10), T = 8.

Peter Lauremberger (Petrus Laurembergus). Institutiones Arithmeticæ .... 4th ed., Joh. Lud. Gleditsch, Leipzig, 1698. P. 196, prob. 12.XII. Fox is 60 hare-leaps ahead of a hound. She makes 9 leaps while the hound does 6, but 7 hare-leaps are as long as 3 hound-leaps.

Isaac Newton. Arithmetica Universalis, 1707. ??NYS. English version: Universal Arithmetic, translated by Mr. Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James Maguire and Theaker Wilder; W. Johnston, London, 1769. (De Morgan, in Rara, 652 653, says there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of 1720 and 1728, ??NYS.) Resolution of Arithmetical Questions, Problem V, pp. 180 184. Begins with MR(7/2, 8/3; 59) with second delayed by 1. Then "The same more generally" does O(c/f, d/g) with second having a headstart of distance e and either a headstart or delay of time h and MR(c/f, d/g; e) with second having either a headstart or delay of time h. Then does example: O(13, 1) with second starting distance 90 ahead, but first delayed by 3 days. Then repeats original example from general viewpoint.

Simpson. Algebra. 1745. Section XI (misprinted IX in 1790).

Prob. XIX, p. 89 (1790: prob. XXXI, p. 92). O-(28, -2; 20, 0).

Prob. L, pp. 113-115. MR-(40, -2; 20, 2; 360).

1790: Prob. LXIII, pp. 114 116. MR-(60, -5; 40, 5; 500).

Prob. LI, pp. 115-116 (1790: prob. LXIV, pp. 116 117). Overtaking. 8 + 12 + ... + (4+4n) = 1 + 4 + ... + n².

Alexis-Claude Clairaut. Élémens d'algèbre. 1746. ??NYS -- cited by Tom Henley. Discusses the relation between overtaking and meeting problems and the use of negative rates or negative headstarts to make them algebraically the same.

Les Amusemens. 1749.

Prob. 113, p. 255. O-(2½, ½; 7, 0).

Prob. 114, p. 256. O-(5, 1; 2, 0).

Prob. 115, p. 257. Meeting: 4n + n+2 + n+4 + n+6 + n+8 = 104.

Prob. 116, p. 258. O-(10, 5; 30, 0).

Prob. 117, p. 259. Cat and mouse, O-(3, 5), D = 23.

Euler. Algebra. 1770. I.IV.III: Questions for practice.

No. 16, p. 205. Privateer and prey, O-(18, 20), D = 8. [The numbers 8 and 18 are interchanged in the text.]

No. 25, p. 206. Hare 50 hare leaps ahead of greyhound. Hare makes 4 leaps to greyhound's 3, but hare's leaps are only ⅔ as long.

Vyse. Tutor's Guide. 1771?

Prob. 57, 1793: p. 68; 1799: p. 74 & Key p. 99. O-(22, 32), T = 4. Misprint in solution.

Prob. 60, 1793: p. 69; 1799: p. 74 & Key P. 100. MR-(x, x+2½; 135) with meeting after 8.

Prob. 65, 1793: p. 69; 1799: p. 75 & Key p. 101. MR-(2, 3; 170) with second delayed by 8.

Prob. 67, 1793: p. 70; 1799: p. 75 & Key p. 102. Hound & hare, O-(21, 15), D = 96. The actual rates are not given, only their ratio, so one can determine where the hare is caught, but not when.

Prob. 12, 1793: p. 78; 1799: p. 84 & Key p. 109. A and B start to circle a wood of circumference 135, starting at opposite sides and going in the same direction. A goes 11/2, B goes 17/3. When do they meet? O (11/2, 17/3), D = 67½.

Prob. 13, 1793: p. 186; 1799: p. 198 & Key p. 241. Two rowers who can row at 5 set out towards each other at points 34 apart on a river flowing 2½. Though this appears to belong in Section 10.G, it is simply MR-(2½, 7½; 34).

Prob. 6, 1793: p. 189; 1799: p. 201 & Key p. 245. Hare starts 5 rods and 34 sec before a greyhound and goes at 12 mph, while the hound goes 20 mph. Using 320 rods to the mile, this is O-(16/15, 16/9) where the slower has a headstart of T = 34 and D = 5.

Dodson. Math. Repository. 1775.

P. 1, Quest. I. MR-(8, 7; 150)

P. 2, Quest. V. O-(30, 42), T = 4.

P. 25, Quest LXIV. Same as Euler's no. 25.

P. 57, Quest. CX. MR-(7/2, 8/3; 59) with second starting one unit of time later.

Pp. 58-59, Quest CXI. General solution of O-(a, b) with one having a headstart and starting before or after the other. Does O-(10/3, 5/2), D = -59, T = 4; then O (10/3, 5/2), D = -59, T = -4.

P. 178, Quest XXIV. O-(8, 0; 1, 2).

P. 179, Quest XXVI. MR-(40, -2; 20, 2; 360).

P. 183, Quest. XXX. 8 + 12 + 16 + ... = 1 + 4 + 9 + ...

Pp. 191-192, Quest XLII. 1 + 2 + 3 + ... + 1 + 8 + 27 + ... = 462.

Charles Hutton. A Complete Treatise on Practical Arithmetic and Book-keeping. Op. cit. in 7.G.2. [c1780?]

1804: prob. 68, p. 139. Hare starts 40 yards before a hound and goes at 10 mph. The hound doesn't see the hare for 40 seconds, and then goes at 18 mph.

1804: prob. 69, p. 139. Exeter is 130 miles from London. A sets out from Exeter at 8  am going 3 mph; B sets out from London at 4 pm going 4 mph. Where do they meet?